Question

Use mathematical induction to prove the statement is true for all positive integers n.

The integer n3 + 2n is divisible by 3 for every positive integer n.

Help me please

Answer 1

Answer on Question #42499– Math – Abstract Algebra

Question:

Use mathematical induction to prove the statement is true for all positive integers $n$.

The integer $n3 + 2n$ is divisible by 3 for every positive integer $n$.

Solution:

**Basis**: Show that the statement holds for $n = 1$.

$$n^3 + 2n = 1^3 + 2 * 1 = 3.$$

3 is divisible by 3, so the statement holds for $n = 1$.

**Inductive step**: Show that if the statement holds for $n = k$, then also the statement holds for $n = k + 1$.

Assume that $k^3 + 2k$ is divisible by 3 (for some unspecified value of $k$). It must then be shown that $(k + 1)^3 + 2(k + 1)$ is divisible by 3 too, that is:

$$\begin{array}{l} (k + 1)^3 + 2(k + 1) = k^3 + 3k^2 + 3k + 1 + 2k + 2 = k^3 + 2k + 3k^2 + 3k + 3 \\ = (k^3 + 2k) + 3k^2 + 3k + 3 = (k^3 + 2k) + 3(k^2 + k + 1) \end{array}$$

It can be easily seen that $(k^3 + 2k) + 3(k^2 + k + 1)$ is divisible by 3. It is because the first term $(k^3 + 2k)$ is divisible by 3 according to the inductive step. And the second term is divisible by 3, because it is a product of 3 and some integer number.

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